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8.5: Pressure Dependence of Gibbs Energy

  • Page ID
    199222
  • The pressure and temperature dependence of \(G\) is also easy to describe. The best starting place is the definition of \(G\).

    \[G = U + pV -TS \label{eq1} \]

    Taking the total differential of \(G\) yields

    \[dG = dU + pdV – pdV + Vdp – TdS – SdT\]

    The differential can be simplified by substituting the combined first and second law statement for \(dU\) (consider a reversible process and \(pV\) work only).

    \[dG = \cancel{TdS} \bcancel{– pdV} + \bcancel{pdV} + Vdp – \cancel{TdS} – SdT\]

    Canceling the \(TdS\) and \(pdV\) terms leaves

    \[dG = V\,dp – S\,dT \label{Total1}\]

    This suggests that the natural variables of \(G\) are \(p\) and \(T\). So the total differential \(dG\) can also be expressed

    \[ dG = \left( \dfrac{\partial G}{\partial p} \right)_T dp + \left( \dfrac{\partial G}{\partial T} \right)_p dT \label{Total2}\]

    And by inspection of Equations \ref{Total1} and \ref{Total2}, it is clear that

    \[\left( \dfrac{\partial G}{\partial p} \right)_T = V\]

    and

    \[ \left( \dfrac{\partial G}{\partial T} \right)_p = -S\]

    It is also clear that the Maxwell relation on \(G\) is given by

    \[\left( \dfrac{\partial V}{\partial T} \right)_p = \left( \dfrac{\partial S}{\partial p} \right)_T\]

    which is an extraordinarily useful relationship, since one of the terms is expressible entirely in terms of measurable quantities!

    \[\left( \dfrac{\partial V}{\partial T} \right)_p = V\alpha\]

    The pressure dependence of \(G\) is given by the pressure derivative at constant temperature

    \[\left( \dfrac{\partial G}{\partial p} \right)_T = V \label{Max2}\]

    which is simply the molar volume. For a fairly incompressible substance (such as a liquid or a solid) the molar volume will be essentially constant over a modest pressure range.

    Example \(\PageIndex{1}\): Gold under Pressure

    The density of gold is 19.32 g/cm3. Calculate \(\Delta G\) for a 1.00 g sample of gold when the pressure on it is increased from 1.00 atm to 2.00 atm.

    Solution:

    The change in the Gibbs function due to an isothermal change in pressure can be expressed as

    \[ \Delta G =\int_{p_1}^{p_2} \left( \dfrac{\partial G}{\partial p} \right)_T dp\]

    And since substituting Equation \ref{Max2}, results in

    \[ \Delta G =\int_{p_1}^{p_2} V dp\]

    Assuming that the molar volume is independent or pressure over the stated pressure range, \(\Delta G\) becomes

    \[\Delta G = V(p_2-p_1)\]

    So, the molar change in the Gibbs function can be calculated by substituting the relevant values.

    \[ \begin{align} \Delta G & = \left( \dfrac{197.0\, g}{mol} \times \dfrac{1\,}{19.32\,g} \times \dfrac{1\,L}{1000\,cm^3} \right) (2.00 \,atm -1.00 \,atm) \underbrace{ \left(\dfrac{8.315 \,J}{0.08206\, atm\,L}\right)}_{\text{conversion unit}}\\ &= 1.033\,J \end{align}\]

    Contributors

    • Patrick E. Fleming (Department of Chemistry and Biochemistry; California State University, East Bay)